Problem: Consider the parametric curve: $\begin{aligned} x&=-3t^{4} \\\\ y&=\cos^3(t) \end{aligned}$ Which integral gives the arc length of the curve over the interval from $t=8$ to $t=13$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_{8}^{13} \sqrt{-12t^3-3\cos^2(t)\sin(t)}\,dt$ (Choice B) B $\int_{8}^{13} \sqrt{9t^8+\cos^6(t)}\,dt$ (Choice C) C $\int_{8}^{13} \sqrt{-3t^{4}+\cos^3(t)}\,dt$ (Choice D) D $\int_{8}^{13} \sqrt{144t^6+9\cos^4(t)\sin^2(t)}\,dt$
Answer: This is the formula for the arc length of a parametric curve over the interval $[a, b]$ : $\int_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\, dt$ [Where does this formula come from?] Let's find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{d}{dt}\left[-3t^{4}\right] \\\\ &=-12t^3 \\\\\\ \dfrac{dy}{dt}&=\dfrac{d}{dt}\left[\cos^3(t)\right] \\\\ &=3\cos^2(t)(-\sin(t)) \\\\ &=-3\cos^2(t)\sin(t) \end{aligned}$ Now we can find the expression for the arc length: $\begin{aligned} &\phantom{=}\int_{8}^{13} \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \\\\ &=\int_{8}^{13} \sqrt{\left(-12t^3\right)^2+\left(-3\cos^2(t)\sin(t)\right)^2}\,dt \\\\ &=\int_{8}^{13} \sqrt{144t^6+9\cos^4(t)\sin^2(t)}\,dt \end{aligned}$ In conclusion, this integral gives the arc length of the curve over the interval from $t=8$ to $t=13$ : $\int_{8}^{13} \sqrt{144t^6+9\cos^4(t)\sin^2(t)}\,dt$